Problem: Simplify; express your answer in exponential form. Assume $z\neq 0, q\neq 0$. $\dfrac{{(z^{-4})^{3}}}{{(zq^{-3})^{-1}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${z^{-4}}$ to the exponent ${3}$ . Now ${-4 \times 3 = -12}$ , so ${(z^{-4})^{3} = z^{-12}}$ In the denominator, we can use the distributive property of exponents. ${(zq^{-3})^{-1} = (z)^{-1}(q^{-3})^{-1}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(z^{-4})^{3}}}{{(zq^{-3})^{-1}}} = \dfrac{{z^{-12}}}{{z^{-1}q^{3}}}$ Break up the equation by variable and simplify. $\dfrac{{z^{-12}}}{{z^{-1}q^{3}}} = \dfrac{{z^{-12}}}{{z^{-1}}} \cdot \dfrac{{1}}{{q^{3}}} = z^{{-12} - {(-1)}} \cdot q^{- {3}} = z^{-11}q^{-3}$.